Problem from Herstein's book (section 2.12, problem 16) If $G$ is a finite group and its $p$-Sylow subgroup $P$ lies in the center of $G$, prove that there exists a normal subgroup $N$ of $G$ with $P\cap N=\{e\}$ and $PN=G$.
Here I have found that this statement follows from Schur-Zassenhaus theorem but I would like to see more simple approach and I would be very grateful if anyone can show how to prove it without using any additional theorems. 
 A: Suppose the order of $G$ is $p^km$ and $P$ is the central $p-$ Sylow subroup. Let us index the (right) cosets of $G$ over $P$ by the integers $[1 \ldots m]$ and choose a representative $r_i$ for each coset. Then each element $g \in G$ can be uniquely written as $g = r_ip_g$ for some $i$ and $p_g \in P$. We will have a closer look at the action of the element $g \in G$ on the set $R$ of representatives. Let us denote the (right) action of $g$ on $r_i$ by $r_ig = r_jg_{i,j}$ where $g$ maps the coset with index $i$ in the coset with index $j$, and where $g_{i,j} \in P$. This allows us to associate a "matrix" $g_{i,j}$ with $g$. It is not hard to see that this "matrix" has the form of a permutation matrix (since $g$ permutes the indices of the cosets) where the ones are replaced with elements of $P$. Suppose another element $h \in G$ acts on $r_j$ by $r_jh = r_kh_{j,k}$ then the composition of these actions (which is also the action of the product $gh$) gives : $r_i \stackrel{g}{\rightarrow} r_jg_{i,j} \stackrel{h}{\rightarrow}r_kg_{i,j}h_{j,k}$ (note that we have implicitly used the fact that $P$ was central). But the action of $gh$ on $r_i$ is by definition $r_k(gh)_{i,k}$ so that $(gh)_{i,k} = g_{i,j}h_{j,k} $, which shows that the "matrix" associated with $gh$ is the product of the matrices associated with $g$ and $h$ respectively. This shows that there is an homomorphism from $G$ to the group generated by the matrices under consideration. It should be noted that the elements of $P$ correspond to scalar matrices. The composition of this homomorphism with the determinant function gives us an homomorphism $G \rightarrow P$ (I suppose that this is the "transfert" function). The intersection of $P$ with the kernel of this homomorphism asks us for the scalar matrices with determinant one, but there is only one since $p \mapsto p ^m$ is an automorphism of $P$. The kernel is the group asked for in the question.
One could argue that this is all very naive, simplistic and not mathematical: the zero entries of the matrices are not elements of $P$, and how are the matrices multiplied if one can not "add"  elements of a (multiplicative) group. This can be rendered in a mathematical rigorous way by introducing  the structure of the module $M$ over the group ring $\mathcal{P}$ constructed as $ M =\bigoplus_{i=1}^{m} \mathcal{P}$ the direct sum of $m$ identical copies of the group ring $\mathcal{P}$. The matrices then represent homomorphisms of the module $M$ and the entries of these matrices are the elements $g_{i,j}$, seen as embedded elements of $P$ in the group ring $\mathcal{P}$.

Example: let G be the group with generators $\{a,b,c\}$ with relators $a^5=1, b^2=1, c^3=1, (bc)^2=1, \{a,b\}=1, \{a,c\}=1$ then $P$ is the unique 5-Sylow subgroup of $G$ generated by $a$. The set of representatives of the cosets can then be chosen as $ a, ba^2, ca^2, c^{-1}a^{-2}b, cb, c^{-1}a^{-1}$. The matrix associated with $b$ is then: 
  $$\begin{pmatrix} 
0&a^{-1}&0&0&0&0\\
a&0&0&0&0&0\\ 
0&0&0&0&a^2&0\\ 
0&0&0&0&0&a^{-1}\\ 
0&0&a^{-2}&0&0&0\\ 
0&0&0&a&0&0\\ 
\end{pmatrix}$$
  the matrix associated with $c$ is: 
  $$\begin{pmatrix} 
0&0&a^{-1}&0&0&0\\
0&0&0&a^{-1}&0&0\\ 
0&0&0&0&0&a^{-2}\\ 
0&0&0&0&a^{-2}&0\\ 
0&a^{-2}&0&0&0&0\\ 
a^{-2}&0&0&0&0&0\\ 
\end{pmatrix}$$ and the matrix associated with $bc$ is:
  $$\begin{pmatrix} 
0&0&0&a^{-1}&0&0\\
0&0&1&0&0&0\\ 
0&1&0&0&0&0\\ 
a^2&0&0&0&0&0\\ 
0&0&0&0&0&a\\ 
0&0&0&0&a^{-1}&0\\ 
\end{pmatrix}$$ which is precisely the product of the two preceding matrices.

